Higher January 2019 Paper 2 Q12
12
(a) Simplify \(n^0\) (1)
(b) Simplify \((3x^2y^5)^3\) (2)
(c) Factorise fully \(2e^2 - 18\) (2)
(d) Make \(r\) the subject of \(m = \sqrt{\dfrac{6a + r}{5r}}\) (4)
| Scheme | Marks |
|---|---|
| 1 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| \(27x^6y^{15}\) | B2 |
| (2) |
Notes
B2: If not B2 then B1 for any two correct terms in a product
| Scheme | Marks |
|---|---|
| \(2(e^2 - 9)\) or \((2e - 6)(e + 3)\) or \((e - 3)(2e + 6)\) | M1 |
| \(2(e - 3)(e + 3)\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(m^2 = \dfrac{6a + r}{5r}\) | M1 |
| \(m^2 \times 5r = 6a + r\) | M1 |
| \(5rm^2 - r = 6a\) | M1 |
| \(r = \dfrac{6a}{5m^2 - 1}\) | A1 |
| (4) | |
| (9 marks) |
Notes
A1: or for \(r = \dfrac{-6a}{1 - 5m^2}\) oe
NB: to award A1 we must see \(r = \dfrac{6a}{5m^2 - 1}\) in working if \(\dfrac{6a}{5m^2 - 1}\) alone is given as answer